integral from-1 to-2 of (16-x^3)/(x^3)

解答

\int_{- 1}^{- 2} \frac{16 - x ^{3}}{x ^{3}} d x

7

求解步骤

\int_{- 1}^{- 2} \frac{16 - x ^{3}}{x ^{3}} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{- 1} \frac{16 - x ^{3}}{x ^{3}} d x

乘开

\frac{16 - x ^{3}}{x ^{3}} : \frac{16}{x ^{3}} - 1

= - \int_{- 2}^{- 1} \frac{16}{x ^{3}} - 1 d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int_{- 2}^{- 1} \frac{16}{x ^{3}} d x - \int_{- 2}^{- 1} 1 d x)

\int_{- 2}^{- 1} \frac{16}{x ^{3}} d x = - 6

\int_{- 2}^{- 1} 1 d x = 1

= - (- 6 - 1)

化简

= 7

\int 5 sin^{3} (5 x) cos^{3} (5 x) d x

\frac{\partial}{\partial x} (x^{2} y + e^{x y})

\int_{0}^{4.5} 90 x d x

\frac{d}{d x} (1 + 2 tan (x))

x \to 0 lim (x^{2} - 1)